CODEWITHSUNDEEP
c++binarybit-manipulationbitwise-operators
Astronautilus
Astronautilus

Reputation: 83

How to split an unsigned long int (32 bit) into 8 nibbles?

I am sorry if my question is confusing but here is the example of what I want to do,

lets say I have an unsigned long int = 1265985549 in binary I can write this as 01001011011101010110100000001101

now I want to split this binary 32 bit number into 4 bits like this and work separately on those 4 bits

0100 1011 0111 0101 0110 1000 0000 1101

any help would be appreciated.

Upvotes: 3

Views: 3376

Answers (3)

user555045
user555045

Reputation: 64913

Based on the comment explaining the actual use for this, here's an other way to count how many nibbles have an odd parity: (not tested)

; compute parities of nibbles
x ^= x >> 2;
x ^= x >> 1;
x &= 0x11111111;
; add the parities
x = (x + (x >> 4)) & 0x0F0F0F0F;
int count = x * 0x01010101 >> 24;

The first part is just a regular "xor all the bits" type of parity calculation (where "all bits" refers to all the bits in a nibble, not in the entire integer), the second part is based on this bitcount algorithm, skipping some steps that are unnecessary because certain bits are always zero and so don't have to be added.

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727047

You can get a 4-bit nibble at position k using bit operations, like this:

uint32_t nibble(uint32_t val, int k) {
    return (val >> (4*k)) & 0x0F;
}

Now you can get the individual nibbles in a loop, like this:

uint32_t val = 1265985549;
for (int k = 0; k != 8 ; k++) {
    uint32_t n = nibble(val, k);
    cout << n << endl;
}

Demo on ideone.

Upvotes: 6

sp2danny
sp2danny

Reputation: 7687

short nibble0 = (i >>  0) & 15;
short nibble1 = (i >>  4) & 15;
short nibble2 = (i >>  8) & 15;
short nibble3 = (i >> 12) & 15;

etc

Upvotes: 4

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